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My question is similar to this one here but a bit different.

I want to know the result of the following operation. I'm familiar with Leibnitz' rule but can't see how to extend it to this case. If it helps, assume that $f$ is a joint density such that at the limits $f(\pm \infty,y)=0,\forall y$. But I'd really like to know the entire formula. $$ \frac{\partial}{\partial x} \left( \int_{-\infty}^\infty \int_{-\infty}^{g(x)} f(x,y)dydx \right) = ?$$

Thanks!

edit: mmh, regarding the second answer below. that makes kind of sense. I need to be a bit more precise here. I'm trying to understand this equation from a paper. Sorry about the notation but that's what they use.

$$ P = \int_{x = -\infty}^\infty \int_{y = -\infty}^{x + a} dF(x,y) \\ = \int_{-\infty}^\infty \frac{\partial F(x,x+a)}{\partial x} dx$$ where $F$ is a joint cdf (so that $dF$ is a joint pdf). Maybe what I wrote above is actually irrelevant at best (and misleading at worst).

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$\int_{-\infty}^\infty \int_{-\infty}^{g(x)} f(x,y)\ dy\ dx$ does not depend on $x$, since $x$ is "integrated out". So the derivative with respect to $x$ is $0$.

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  • $\begingroup$ hey! thanks for your answer. actually what you said made me reformulate the question. do you mind giving another look? thanks a ton. $\endgroup$ – Florian Oswald Feb 7 '13 at 20:49
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For clarity it is better to use a different variable instead of $x$ either under the inner integral, or in the upper limit and differentiation operator. Say $$I=\frac{\partial}{\partial x}\int_{-\infty}^{+\infty}\int_{-\infty}^{g(x)}f(t,y) \, dy \, dt$$ Then the Leibnitz rule applies provided you can justify the interchanging of integration and differentiation.

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  • $\begingroup$ hi guys, thanks for the answer. apologies for the notation, trying to relate that to a paper. I edited the question above a little, hopefully it got clearer. $\endgroup$ – Florian Oswald Feb 7 '13 at 20:50

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